$\newcommand{\w}{\mathscr{W}}$I cite this text on dynamics, from chapter $2$ in the early pages, mainly section $2.4$. They use the convention that $\Bbb N$ does not include $0$. Some context:
For $k\in\Bbb N$, define $L=\{0,1,\cdots,k-1\}$, the $k$-alphabet, its elements the letters, endowed with the discrete topology and define the product topological (and compact) language, the space of all words: $$\w_k^+:=L^{\Bbb N_0}$$Define the continuous shift-dynamic $\tau:\w_k^+\to\w_k^+,\,(x_n)_{n\in\Bbb N_0}\mapsto(x_{n+1})_{n\in\Bbb N_0}$.
Define an $n$-block to be an element of $L^n$. For any arbitrary $B\subseteq\bigcup_{n\in\Bbb N_0}L^n$, call it the collection of excluded blocks and define: $$\w_k^B=\{x\in\w_k^+:\text{ no block of $x$ is contained in $B$}\}$$Such a set is $\tau$-invariant.
It is shown in the book that any subsystem of $(\w_k^+;\tau)$ (a subsystem is a nonempty closed $\tau$-invariant subspace with dynamic the restriction of $\tau$ to that subspace) $(F;\tau)$ must satisfy $F=\w_k^{B_F}$ for some such family of excluded blocks $B_F$. $(F;\tau)$ is a subshift of the finite type if $B_F$ is finite, of order $n$ if $|B|=n\in\Bbb N_0$. By shift-invariance one may wlog extend all the blocks, in all possible ways, in $B_F$ to have length $n$.
The following is the part that confuses me:
For a subshift $(F;\tau)$ of finite type, order $2$, associate the transition matrix $A:=(a_{ij})^{k-1}_{i,j=0}$ by: $$a_{ij}=\begin{cases}1&(i,j)\notin B_F\\0&(i,j)\in B_F\end{cases}$$
Apparently this is indeed a "transition matrix", but I don't see why.
In Markov chains (which I only very cursorily studied) a transition matrix $A$ defines the sequence by $Ax_n=x_{n+1}$. In analogy, the transition matrix of $F$ is apparently going to perform the shift action, but I don't see how that is since, in the context of Markov chains $x_{n+1}:=Ax_n$ by definition, but in $\w_k^+$ the sequence $(x_0,x_1,\cdots)$ is given: then $A$ has no knowledge of $x_{n+1}$, so we must prove somehow that $A(x)=\tau(x)$.
I am not sure in what sense $A$ would take the input however. It is $k\times k$ square, so we expect a $k$-vector as input. My only thought is that the input is supposed to be a one-hot encoding of a single letter in the word, with $0$s everywhere except for a $1$ at the position indicating the value $0,1,\cdots,k-1$, but then $A$ might send this input to a vector which is not one-hot, so I'm confused. It could be the case that $A$ accepts the next $k$ letters in the word as input, perhaps convolving over the word $x\in F$, but this doesn't make sense to me either.
As a concrete example, consider $\w_3^+$ and the subsystem obtained by excluding the blocks $(1,2),(0,0),(2,0)$. The transition associated to this order-$2$ subshift is: $$A=\begin{bmatrix}0&1&1\\1&1&0\\0&1&1\end{bmatrix}$$Consider a word $x$ in this subsystem beginning with $(0,1,1,0,2,2,1,\cdots)$. In what sense can we say that $\tau(x)=(1,1,0,2,2,1,\cdots)$ is represented by the action of $A$? I am missing something very basic here I think.
I think the matrix only describes $F$, and doesn't really have anything to do with the shift operator $\tau$.
An order $2$ subshift is completely specified by the set $B_F$ of excluded blocks of length $2$, and one way to encode this is by having binary indicators for each possible block of length $2$ not in $B_F$, and you can arrange these indicators in a matrix. To relate this back fo $F$, the entry $a_{ij}$ can be interpreted as an indicator for the event
$$\text{The block $(i,j)$ appears in some word of $F$.}$$
This may seem weird if you are thinking about Markov transition matrices, since here the entries are binary instead of probabilities, but it may be useful to think about the state space as your alphabet $L$, and then $a_{ij}$ being an indicator for $$\text{The letter $i$ followed by the letter $j$ in some word of $F$.}$$ Then, similar to Markov transition matrices, there is a natural interpretation for the entries of $A^n$, since the $(i,j)$ entry of $A^n$ is an indicator for $$\text{The letter $i$ is followed by $n-1$ letters and then the letter $j$ in some word of $F$.}$$ This is essentially Lemma 2.40.
If you insist on thinking about $A$ in terms of multiplication with a vector, you can let $v$ be a vector of $1$s and $0$s representing some subset $L'$ of letters $L$ (specifically, $v_i = \mathbf{1}_{i \in L'}$), and then $Av$ will also be a vector of $1$s and $0$s, representing the letters that appear immediately after a letter of $L'$ in some word of $F$.